Suppsoe that $$c = \frac{(a+b)(a+b+1)}{2} + b$$
Now $c$ is given - how does one find satisfying $a, b$?
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https://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function – Mar 13 '13 at 12:54
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You can set $d=a+b$ and solve the resulting equation $d^2+3d-2(a+c)=0$ in $d$. This gives a dependency $d(a)$ and hence $b(a)$. – Nikolaj-K Mar 13 '13 at 12:58
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Presumably, $a,b,c$ are natural numbers. – Thomas Andrews Mar 13 '13 at 13:15
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@NickKidman Where did the $3$ come from? – Thomas Andrews Mar 13 '13 at 13:17
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@ThomasAndrews: $\frac{d(d+1)}{2}+d+x=\frac{3}{2}d+y$. – Nikolaj-K Mar 13 '13 at 13:53
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I've edited the title to reflect the domain where $a,b,c$ belong. As you question is tagged as "number-theory", I think, as Thomas Andrews suggests, you mean $a,b,c$ are natural numbers. Please roll back if I am wrong. – user1551 Mar 13 '13 at 14:04
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Ah, hadn't noted the $b$ at the end became an $a$. Yeah, you can just have kept $d^2+d-2(c-b)$. @NickKidman – Thomas Andrews Mar 13 '13 at 14:11
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Assuming you have the condition that $a,b,c$ are natural numbers.
Let $T_n=\frac{n(n+1)}{2}$ be the $n$th triangular number.
Let $T_m$ be the largest triangular number less than or equal to $c$.
Show that $c-T_m\leq m$
Let $a=c-T_m$ and $b=m-a$.
You can actually get an explicit formula for $m$, namely: $$m=\left\lfloor \frac{-1+\sqrt{8c+1}}{2}\right\rfloor$$
Thomas Andrews
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the equation:
$c=\frac{(a+b)(a+b+1)}{2}+b$
Solutions can be written:
$a=\frac{q(q\pm1)}{2}-c-1$
$b=c-\frac{q(q\mp1)}{2}$
$q$ - what some integer.
individ
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